Practice Problems on Group Theory 10.1 Q16 Let H = {0, 3, 6, 9} be a subset of Z12 (the integers modulo 12 under addition). Prove that H is a subgroup of Z12 . Solution: To prove H ≤ Z12 , verify the subgroup criteria: 1. Non-empty: H contains 0 (identity element). 2. Closure: For a, b ∈ H, a + b (mod 12) ∈ H. Check pairwise sums: 3 + 3 = 6, 3 + 6 = 9, 3 + 9 = 12 ≡ 0 (mod 12), etc. 3. Inverses: For a ∈ H, −a (mod 12) ∈ H. For example, −3 ≡ 9 (mod 12) ∈ H. Since all conditions are satisfied, H is a subgroup of Z12 . 10.3 Q1 Consider the homomorphism ϕ : Z → Z6 defined by ϕ(n) = n (mod 6). Determine the kernel of ϕ. Solution: The kernel of a homomorphism is ker(ϕ) = {n ∈ Z | ϕ(n) = 0}. Solve n ≡ 0 (mod 6), i.e., n = 6k for k ∈ Z. Thus, ker(ϕ) = 6Z, the set of integers divisible by 6. 1 10.3 Q4 Show that the symmetric group S3 is solvable. Solution: The derived series for S3 is: (0) S3 = S3 , (1) S3 = [S3 , S3 ] = A3 = {e, (123), (132)}, (2) S3 = [A3 , A3 ] = {e}. Since the derived series terminates at the trivial group, S3 is solvable. 10.5 Q1 Let G = Z6 act on X = {0, 1, 2, 3, 4, 5} by addition modulo 6. Find the size of the orbit and stabilizer of the element x = 0. Solution: • Orbit: Orb(0) = {g · 0 | g ∈ Z6 } = {0}. Hence, |Orb(0)| = 1. • Stabilizer: Stab(0) = {g ∈ Z6 | g + 0 ≡ 0 (mod 6)} = Z6 . Thus, |Stab(0)| = 6. By the orbit-stabilizer theorem, |G| = |Orb(0)|·|Stab(0)|, verified as 6 = 1·6. 10.5 Q2 Prove that a group of order 20 has a normal Sylow 5-subgroup. Solution: • Sylow 5-Subgroups: The number of Sylow 5-subgroups satisfies n5 ≡ 1 (mod 5) and divides 20/5 = 4. Possible values are n5 = 1 or 4. • If n5 = 1, the Sylow 5-subgroup is unique and hence normal. 2 • If n5 = 4, a contradiction arises because G would then have 4 · 4 = 16 elements of order 5, leaving no room for other elements. Thus, n5 = 1, and the Sylow 5-subgroup is normal. 3